[jackie11]

Hi there

Please could someone take a loko at the below and for question 4, tell me why I am getting a slightly different answer to whats in the book, and for question 5, explain how to do this as I have completely gone off the rails lol

I am really struggling with these types of questions, so if anyone can give me a little Differential equations for dummies guide or point me in the direction of one, I would be most grateful, cheers;


Question 4:



Question 5:





Many thanks

[Mr M]

First one: How on Earth are you getting exponentials? Separate the variables and integrate!

[mathz]

first one:

integral 1/h^1/2 dh = integral k dt

where do you get exp from?

[Mr M]

Second one is no better.



Factorise the denominator of the integrand and split it into partial fractions.

[mathz]

1/(x(x-1) does integrate to ln(x(x-1)

you need to split it into partial fractions.

[mathz]

lol,there is an echo is this thread

[TenOfThem]

And I can only echo too

OP you seem to have misunderstood this topic completely

You do not even begin correctly

[Mr M]

(Original post by TenOfThem)
And I can only echo too

OP you seem to have misunderstood this topic completely

You do not even begin correctly

I ignored that as the negative sign can be incorporated within the constant of proportionality. Think the other problems are more glaring!

[TenOfThem]

(Original post by Mr M)
Think the other problems are more glaring!

Agreed

[jackie11]

ok thank you all for your replies.

I have managed to solve question 5, but am still having trouble with question 4.

With question 4, is this going in the right direction;

2√h = -kt + c

4h = (-kt)² + c

h = 1/4(-kt)² + c

???

Then does, c = 200?

[Mr M]

(Original post by jackie11)
ok thank you all for your replies.

I have managed to solve question 5, but am still having trouble with question 4.

With question 4, is this going in the right direction;

2√h = -kt + c

4h = (-kt)² + c

h = 1/4(-kt)² + c

???

Then does, c = 200?

[Quip]

(Original post by jackie11)
ok thank you all for your replies.

I have managed to solve question 5, but am still having trouble with question 4.

With question 4, is this going in the right direction;

2√h = -kt + c

4h = (-kt)² + c

h = 1/4(-kt)² + c

???

Then does, c = 200?

Definitely going in the right direction, but I'm not convinced on your workings between lines 1 and 2 as you should square all of the right hand side as well. I know that a constant squared is just another constant, but you should also get -2ckt which you haven't got. Therefore, I would just substitute some values into the top line to find values for k and c. I think that works out fine.

[jackie11]

(Original post by Quip)
Definitely going in the right direction, but I'm not convinced on your workings between lines 1 and 2 as you should square all of the right hand side as well. I know that a constant squared is just another constant, but you should also get -2ckt which you haven't got. Therefore, I would just substitute some values into the top line to find values for k and c. I think that works out fine.

ok how about this;

2√h = -kt + c

when h = 200, t = 0

2√200 = c

when h = 128, t = 5

2√128 = -5k + 2√200
2√128 - 2√200 = -5k
-5.656 = -5k
5.656 = 5k
1.13 = k

[Quip]

(Original post by jackie11)
ok how about this;

2√h = -kt + c

when h = 200, t = 0

2√200 = c

when h = 128, t = 5

2√128 = -5k + 2√200
2√128 - 2√200 = -5k
-5.656 = -5k
5.656 = 5k
1.13 = k

If this is still relevant, yes that's right (I think), but it may be more useful to keep k as a surd though.